If cos (alpha+beta) = 4 / 5, sin (alpha-beta) = 5 /13 and (alpha) and (beta) lies between 0 and pi then tan 2 alpha is
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cos (a + b ) = 4/5 and sin(a-b) = 5/13 , 0 < a, b , <π
0 < a < π
0 < b < π
tan2a = ?
cos (a+b) = 4/5
sin (a+b) = 3/5
a + b = sin ⁻¹ (3/5) ...............(1)
and
sin(a-b) = 5/13
a -b = sin ⁻¹ (5/ 13 ) ............(2)
adding 1 and 2 gives :
2a = sin⁻¹3/5 + sin⁻¹5/13 = sin⁻¹ (3/5 x 12/13 + 5/13 x 4/5 )
From formula :
[ sin⁻¹ a + sin⁻¹ b = sin⁻¹ (a√1 - b² + b√ 1- a² ) ]
2a = sin⁻¹ (36 + 20 /65 ) = sin⁻¹(56/65)
sin2a = 56/65 , cos2a = 33/65
tan2a = 33/65
Now for its sign:
0 < a < π
0 < 2a < 2π
So tan2a can have values : + 33 /56 , - 33/56
Hope my answer is correct.
0 < a < π
0 < b < π
tan2a = ?
cos (a+b) = 4/5
sin (a+b) = 3/5
a + b = sin ⁻¹ (3/5) ...............(1)
and
sin(a-b) = 5/13
a -b = sin ⁻¹ (5/ 13 ) ............(2)
adding 1 and 2 gives :
2a = sin⁻¹3/5 + sin⁻¹5/13 = sin⁻¹ (3/5 x 12/13 + 5/13 x 4/5 )
From formula :
[ sin⁻¹ a + sin⁻¹ b = sin⁻¹ (a√1 - b² + b√ 1- a² ) ]
2a = sin⁻¹ (36 + 20 /65 ) = sin⁻¹(56/65)
sin2a = 56/65 , cos2a = 33/65
tan2a = 33/65
Now for its sign:
0 < a < π
0 < 2a < 2π
So tan2a can have values : + 33 /56 , - 33/56
Hope my answer is correct.
Answered by
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Answer:
cost (a+b) and sin (a-b) 5/3. 0
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